Let $(X, \mathcal B, \mu)$ be a "good" measure space, e.g. $\mu$ is a positive Radon measure on a locally compact topological space $X$ with Borel $\sigma$-algebra $\mathcal B$. Let $A\subset X$ such that every measurable subset of $A$ has zero measure. Is it true that there is a zero measure set $B$ such that $A\subset B$?
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You are asking whether every set with inner measure $0$ has measure $0$ with respect to the completion measure. The Lebesgue measure, for example, does not have this property, since the usual Vitali set is non-measurable, but has inner measure $0$.
I claim that a ($\sigma$-finite) measure space has your property if and only if every set is measurable with respect to its completion measure.
For the forward direction, suppose that $A$ is any subset of the space $X$. Let $a$ be the inner measure of $A$, the supremum of $\mu(A_0)$ among all measurable $A_0\subset A$ (and we may assume wlog this is finite). By taking a union, it follows that the inner measure is realized, so that there is some measurable $A_0\subset A$ with $\mu(A_0)=a$. It follows that $A-A_0$ has inner measure $0$. By the hypothesis, it follows that $A-A_0\subset B$ for some measurable set $B$ with $\mu(B)=0$, and so $A-A_0$ has measure $0$ with respect to the completion. Consequently, $A$ differs from the measurable set $A_0$ on a completion-measure zero set $A-A_0$, and hence is measurable with respect to the completion measure.
Conversely, suppose that every set is measurable with respect to the completion of $\mu$. Suppose that $A$ has inner measure $0$. By assumption, there is some measurable set $A_0$ such that the symmetric difference $A\triangle A_0\subset A_1$ for some measurable $A_1$ with $\mu(A_1)=0$. It follows that $A_0-A_1$ is a measurable subset of $A$, and hence measure $0$, and so $A\subset B=A_0\cup A_1=(A_0-A_1)\cup A_1$ shows that $A$ is contained in a measure $0$ set $B$, as desired.
In particular, a complete measure has the property if and only if it measures every set.
answered Dec 17, 2010 at 4:57
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2$\begingroup$ On the other hand, if one assumes determinacy (that contradicts choice), then the result is true (at least in any decent space). For example, for the reals we in fact have that if the only Borel subsets of $A$ are measure zero, then $A$ is contained in a Borel set of measure zero. Note that Joel's argument does not use choice after the first paragraph. $\endgroup$ Commented Dec 17, 2010 at 15:40